1. First we show that what we wish to prove for general
n actualy holds for the first n which is realevant to the
problem. In this case it's n=0, and we need to show that
for every x>0 that:

,

which I presume you know is true.

2. Second we show that if what we have to prove is true
when n=k, then it is also true for n=k+1. So assume it
true for n=y, then we have

,

for all .

Now consider:

also:

Now the integrand in [1] is strictly greater than the integrand in
[2] at every point over which the integral is taken. So

which can be rewritten from what we have found above as:

So rearranging and replacing y by x we have:

3. That is we have proven that if what we have to prove for all ,
is true for n=k, then it is true for n=k+1, also we have proven that
it is true for n=0, so by the principle of induction we have proven
it true for all

Hmm, I actually thought of that, but for some reason I thought that was basically just cheating. I had assumed that it would have something to do with how e^x with x<0 would just keep approaching 0. However, I can't come up with anything that satisfies me, so I will go with your more experienced suggestion. Thanks again.